Question

A solution of $0.1M$ ${Cl}^{-}$, $0.1M$ ${Br}^{-}$ and $0.1M$ $I^{-}$ spolid $Ag{NO}_3$ is gradually added to this solution. Assuming that the addition of $Ag{NO}_3$ does not change the volume.
Given: $K_{sp}\left(AgCl\right)=1.7\times {10}^{-10}$, $K_{sp}\left(AgBr\right)=5\times {10}^{-13}$ $K_{sp}\left(AgI\right)=8.5\times {10}^{-17}$ What will be the conc. of both ions when the third ion start precipitating.

• A. ${Br}^{-}=2.9\times {10}^{-4}$; $I^{-}=5\times {10}^{-8}$
• B. ${Br}^{-}=5\times {10}^{-8}$; $I^{-}=2.9\times {10}^{-4}$
• C. ${Br}^{-}=5.8\times {10}^{-4}$; $I^{-}=10\times {10}^{-8}$
• D. ${Br}^{-}=10\times {10}^{-4}$; $I^{-}=5.8\times {10}^{-8}$

Answer 1

The correct option is A ${Br}^{-}=2.9\times {10}^{-4}$; $I^{-}=5\times {10}^{-8}$

The solubility product value of $AgCl$ is maximum. Hence, it will precipitate last.
$K_{SP,AgCl}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$ The chloride ion concentration is 0.10 M.
$1.70\times {10}^{-10}=\left[A{g}^{+}\right]\times 0.10$
$\left[A{g}^{+}\right]=1.70\times {10}^{-09}\text{M}$
Hence, when AgCl will start precipitating, the silver ion concentration will be $1.70\times {10}^{-09}\text{M}$
$K_{SP,AgBr}=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$
$5.00\times {10}^{-13}=1.70\times {10}^{-09}\times \left[B{r}^{-}\right]$
$\left[B{r}^{-}\right]=2.94\times {10}^{-04}\text{M}$
Hence, when AgCl will start precipitating, the bromide ion concentration will be $2.94\times {10}^{-04}\text{M}$

$K_{SP,AgI}=\left[A{g}^{+}\right]\left[I^{-}\right]$
$8.50\times {10}^{-17}=1.70\times {10}^{-09}\times \left[I^{-}\right]$
$\left[I^{-}\right]=5.00\times {10}^{-08}\text{M}$
Hence, when AgCl will start precipitating, the iodide ion concentration will be $5.00\times {10}^{-08}\text{M}$